Question

How do you integrate `int x^2 cos x^2 dx` using integration by parts?

Answer 1

`int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C`

Explanation:

When we integrate by parts a function of the form:

`x^nf(x)`

we normally choose `x^n` as the integral part and `f(x)` as the differential part, so that in the resulting integral we have `x^(n-1)`

In this case however

`cos^2xdx`

is not the differential of an «easy» function, so we first reduce the degree of the trigonometric function using the identity:

`cos^2x = (1+cos(2x))/2`

`int x^2cos^2xdx = int x^2(1+cos(2x))/2dx = int x^2/2dx +int x^2/2 cos(2x)dx`

Now we can solve the first integral directly:

`int x^2/2dx = x^3/6`

and the second by parts:

`int x^2/2 cos(2x)dx = 1/4 int x^2 d(sin2x) = (x^2sin(2x))/4 -1/2 int xsin(2x)dx`

and again:

`1/2 int xsin(2x)dx = -1/4 int xd(cos2x) = -(xcos(2x))/4 + 1/4 int cos2xdx = -(xcos(2x))/4 +1/8sin2x +C`

Putting it together:

`int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C`